Equation 3 x - a^3 + 5 x - a^5 + 7 x - a^7 = 0,a > 1 has - Vidyadip

MCQ

Equation $\frac{3}{{x - {a^3}}} + \frac{5}{{x - {a^5}}} + \frac{7}{{x - {a^7}}} = 0,a > 1$ has

✓
Two real and positive roots
B
Two real and negative roots
C
No real roots
D
one positive and other negative roots

✓

Answer

Correct option: A.

Two real and positive roots

a
Equation $3\left(x-a^{5}\right)\left(x-a^{7}\right)+5\left(x-a^{3}\right)\left(x-a^{7}\right)+7\left(x-a^{3}\right)\left(x-a^{5}\right)=0$

${\rm{ Now }}\,\,\,a > 1 \Rightarrow {a^3} < {a^5} < {a^7}$

${\rm{ Let }}f(x) = 3\left( {x - {a^5}} \right)\left( {x - {a^7}} \right) + 5\left( {x - {a^5}} \right)\left( {x - {a^7}} \right) + $

${7\left(x-a^{3}\right)\left(x-a^{5}\right) /\left(x-a^{3}\right)\left(x-a^{3}\right)\left(x-a^{7}\right)} $

$f\left( {{a^{3 - }}} \right) < 0,f\left( {{a^{3 + }}} \right) > 0,f\left( {{a^{5 - }}} \right) < 0$

$f\left( {{a^{5 + }}} \right) > 0,f\left( {{a^{7 - }}} \right) < 0\,\,and\,\,f\left( {{a^{7 + }}} \right) > 0$

$\therefore$ Equation has one root lying in $\left(a^{3}, a^{5}\right)$ and other in $\left(a^{5}, a^{7}\right)$

Now $a>1 \therefore$ Roots are real and positive

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